STEP Maths I, II, III 1997 Solutions

Speleo

Looks like we only have II/5 and III/8 to go then.I know how to do them, only question is if I can find the time.

Reply 43

1997 III, Q5 (finally):

Wlog, the wheel has radius 1. Let the angle turned through be t. If the wheel wasn't rolling, the position of a point on the rim is (-sin t, 1-cos t). As the wheel is rolling, we have to add t to the x coord to get (t-sin t, 1 - cos t).

Distance travelled by point on rim =

\int_0^{2\pi}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt = \int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t} dt

= \int_0^{2\pi}\sqrt{2 - 2 \cos t} dt = 2 \int_0^{2\pi}\sqrt{\frac{1}{2}(1-\cos t)} dt

= 2 \int_0^{2\pi}\sqrt{\sin^2 \frac{t}{2}}\, dt = 2 \int_0^{2\pi} |\sin \frac{t}{2}| dt

(note that we must take the positive root).

This =

2\int_0^{2\pi} \sin \frac{t}{2} dt = -4[ \cos \frac{t}{2}]_0^{2\pi} = 8.

(N.B. Corrections due to ukgea's post below...)

Distance traveled by center = 2 pi. So ratio is

\pi / 4

.

[OK, what have I missed? Far, far too easy for STEP III]

Reply 44

Shouldn't it be

\displaystyle = \int_0^{2\pi} 2\sqrt{\sin^2 \frac{t}{2}}dt = \cdots

which gives the final answer

\pi/16

instead? Other than that, it looks correct. I guess the hard part is realising that the wheel is moving with a speed

\omega r

or something like that. And the fact that the integral will look more intimidating if you don't assume

\omega = r = 1

as you have done. But yes, it seems rather easy.

Edit: Oh now I realise t doesn't mean time, but angle. My integrals have t = time and consequently a lot of

\omega

s floating around. Nifty. Never mind the part about

\omega

then.

Reply 45

1997 III, Q6:

If

x = \cos \theta

then

\frac{d}{d\theta} = \frac{dx}{d\theta}\frac{d}{dx} = -\sin \theta \frac{d}{dx}

.

So

\displaystyle \frac{dy_n}{d\theta} = - \sin \theta \frac{dy_n}{dx}, \frac{d^2y_n}{d\theta^2} = \sin^2 \theta \frac{d^2y_n}{dx^2} - cos\theta \frac{dy_n}{dx} = (1-x^2)\frac{d^2y_n}{dx^2} - x \frac{dy_n}{dx}

So

\displaystyle \frac{d^2y_n}{d\theta^2} + n^2 y_n = 0

as desired.

This has general solution

y_n = A_n \cos n\theta + B_n \sin n\theta

.

y_n(1) = 1 \implies y_n(\cos 0) = 1 \implies A_n = 1

.

To use the 2nd boundary condition, observe

-\cos \theta = \cos(\pi - \theta)

and so we have:

\cos n\theta + B_n \sin n\theta = (-1)^n (\cos n(\pi-\theta) + \sin n(\pi-\theta))

RHS = (-1)^n(\cos n\pi \cos n\theta -\cos n\pi \sin n\theta) = \cos n\theta - B_n \sin n\theta

and so

B_n = 0

.

So

y_n = \cos n \theta

(which is also

\cos(n \arccos x)

). Finally, we need to show:

\cos(n+1)\theta - 2\cos \theta \cos n\theta + cos(n-1)\theta = 0

.
But

\cos(n+1)\theta = \cos n\theta \cos \theta - \sin n \theta \sin \theta, \cos(n-1)\theta = \cos n\theta \cos \theta + \sin n \theta \sin \theta

, hence result.

Reply 46

khaixiang

5

Question 6, STEP III 1997

(1)

\\ \displaystyle \frac{dy_{n}}{dx} = \frac{dy_{n}}{d\theta}\times \frac{d\theta}{dx}\\ =\frac{dy_{n}}{d\theta}\times\frac{1}{-\sin\theta}

Differentiate (1) to get (2):

\displaystyle \frac{d^{2}y_{n}}{dx^2}=\frac{d^{2}y_{n}}{d\theta^2}\times\frac{1}{\sin^2\theta}-\frac{dy_{n}}{d\theta}\times\frac{\cos\theta}{\sin^3\theta}

Substitute (1), (2) and

x=\cos\theta

into

\displaystyle (1-x^2)\frac{d^{2}y_{n}}{dx^2}-x\frac{dy_{n}}{dx}+n^2y_{n}=0

to get

\displaystyle \frac{d^2y}{d\theta^2}+n^2y_{n}=0\text{ as required }

The general solution of this linear homogeneous second order differential equation is

y_{n}(\theta)=A\cos n\theta+B\sin n\theta

We are given y(1)=1, hence after the tranformation of

x=\cos\theta

above, we have the boundary condition y(0)=1 which results in A=1. Furthermore, we are given

y_{n}(x)=(-1)^ny_{n}(-x)

which really means that

y_{n}(x)

is an even function when n is either 0 or even. And that it's an odd function when n is odd. But after the substitution of

x=\cos\theta

,

y_{n}(\theta)

becomes an even function for all n since cosine is an even function.

We have

\\y_{n}(\theta)=A\cos n\theta+B\sin n\theta\\y_{n}(-\theta)=A\cos n\theta-B\sin n\theta

which can only be even for all n when B=0

\therefore y_{n}(\theta)=\cos n\theta

and

y_{n}(x)=\cos (n\arccos x)

for

|x|\leq1

It follows that

y_{0}(x)=\cos 0 =1

and

y_{1}(x)=\cos(\arccos x)=x

as required.

\\y_{n}(x)=\cos (n\arccos x)\\y_{n-1}(x)=\cos\{(n\arccos x)-(\arccos x)\}\\y_{n+1}(x)=\cos\{(n\arccos x)+(\arccos x)\}

Upon expanding with Trigonometric Addition Formulas and adding

y_{n+1}(x)

and

y_{n-1}(x)

:

\\y_{n+1}(x)+y_{n-1}(x)=2x\cos (n\arccos x)\\y_{n+1}(x)+y_{n-1}(x)=2xy_{n}(x)\\y_{n+1}(x)-2xy_{n}(x)+y_{n-1}(x)=0

as shown.

EDIT: Sorry, didn't know Dfranklin was doing this question

Reply 47

ukgea

Shouldn't it be

\displaystyle = \int_0^{2\pi} 2\sqrt{\sin^2 \frac{t}{2}}dt = \cdots

which gives the final answer

\pi/16

instead? I think you're right about the trig identity (that will teach me to do the question straight into Latex instead of paper), but I don't think it actually affects the final answer; my integral is 4 times more "squashed in x" that it should be, but it repeats 4 times in the interval, so it evens out.

Reply 48

DFranklin

I think you're right about the trig identity (that will teach me to do the question straight into Latex instead of paper), but I don't think it actually affects the final answer; my integral is 4 times more "squashed in x" that it should be, but it repeats 4 times in the interval, so it evens out.

Err, you're right. I forgot about the limits. :smile:

Reply 49

dvs

10

STEP III

Q8.
(I'm using 'q' for alpha, and t for theta.)

(i)
If we just carry out the multiplication, we'll find that the off-diagonals are the same. If we equate either to zero, we get the following equation:
-asin(q)cos(q) + bcos^2(q) - bsin^2(q) + ccos(q)sin(q) = 0
=> (c-a)sin(q)cos(q) + b(cos^2(q) - sin^2(q)) = 0
=> (c-a)sin(2q)/2 + bcos(2q) = 0
=> tan(2q) = 2b/(a-c)
=> q = 1/2 arctan(2b/(a-c))

(ii)
Let's expand the equation of the ellipse:
x^2 + y^2 + 4xy cot(2t) + 4x^2 cot^2(2t) = 1

And also, let's expand the matrix equation:
ax^2 + 2bxy + cy^2 = 1

Now let's equate coefficients:
a = 1 + 4cot^2(2t)
b = 2cot(2t)
c = 1

So we have what A looks like in this case.

If it's diagonal, then:
q = 1/2 arctan(2b/(a-c))
= 1/2 arctan(4cot(2t)/4cot^2(2t))
= 1/2 arctan(1/cot(2t))
= 1/2 arctan(tan(2t))
= t, as required.

Let's go back to the R(-q) A R(q) we found in the first part of the question. Here we need to replace a,b,c with our new values, and q with t. We find, upon simplification, that the non-diagonal elements are (respecitvely, top left and bottom right):
cot^2(t)
tan^2(t)

(iii) What R(-q) A R(q) does is shift the xy-axes into x'y'-axes so that they align with the major and minor axes of the ellipse. Let's write down the new equation of the ellipse (using the matrix equation in part (ii), but using R(-q) A R(q) instead of A; and (x', y') instead of (x, y), to emphasise the coordinate switch):
(x')^2 cot^2(t) + (y')^2 tan^2(t) = 1

Here, the minor axis has length tan(t) and the major axis has length cot(t). Done.
---------------------

I hope it's OK that I left out most of the calculations and simplifications. It would've taken too much time to TeX/type them up.

Reply 50

dvs

10

STEP II

Q5.

If z = ie, then if we equate real and imaginary parts of the equation z = e^w we get:
e^u cos(v) = 0
e^u sin(v) = e

Since e^u > 0, then cos(v) = 0 and thus v = pi/2 + kpi. Whence e^u = e, and so u=1. That is,
w = 1 + i(pi/2 + kpi)

If u is constant, say u=c, then
x + iy = e^c cos(v) + i e^c sin(v)
=> x/e^c = cos(v) and y/e^c = sin(v)

Thus
x^2 + y^2 = e^(2c)

So the corresponding locus in the x-y plane is simply a circle centered at the origin of radius e^c.

On the other hand, if v is constant, say v=k, then
x + iy = e^u cos(k) + i e^(u) sin(k)
=> x/cos(k) = e^(u) = y/sin(k)

Thus
y = tan(k) x

That is, the corresponding locus in the x-y plane is a straight line through the origin of gradient tan(k), i.e. it makes an angle of k with the positive x-axis.

So, for the last bit, the following subsets do the trick:
W_1 = { (u,v) : u in R, 0 < v < pi/2 }
W_2 = { (u,v) : u in R, 2pi < v < 5pi/2 }
W_3 = { (u,v) : u < 0, 0 <= v < 2pi }
W_4 = { (u,v) : u < 0, 2pi <= 0 < 4pi }

(W_3 and W_4 correction due to DFranklin - see post below. Originally I split the set {(u,v) : u<0, v in R} incorrectly.)

Reply 51

I'd just ground through these last two and now I see you've done them!

Did you find any "nice" way of finding the diagonal elements in III/8? I couldn't find one, and did it as a pure trig algebra grind, reducing everything to sin and cos, and it was really really nasty. It didn't seem remotely feasible in a "par" time of 30 minutes.

As far as the complex locus question, I don't think W3, W4 are 1-1.

Don't

(-1, -\pi), (-1, 3\pi), (-1, -\pi), (-1, -3\pi)

all map to -1/e?

I think it works if you instead restrict v by:

W3: u<0, 0<=v<2pi
W4: u<0 2pi<=v<4pi.

(basically any disjoint intervals of length 2pi will work, with slight care about end points).

Or am I missing the point?

Reply 52

dvs

10

I think your W_3 and W_4 work. Restricting v to such intervals of length 2pi will make z go around the circle one revolution. I'll edit my post. :smile:

As for III/8, I did what you did. It was a really long question with lots of opportunities for error. I would have definitely skipped it if I were doing this paper.

Reply 53

insparato

15

Whenever i see these STEP papers i can only do about 1 or 2 questions at the most. I feel as if im lacking something knowledge wise all the time. The ones i can attempt are challenging and like them. But some of the solutions people have posted i havent even come across some of the notation in them! Does STEP imply that people doing these read around what they've been taught?

Reply 54

insparato

Whenever i see these STEP papers i can only do about 1 or 2 questions at the most. I feel as if im lacking something knowledge wise all the time. The ones i can attempt are challenging and like them. But some of the solutions people have posted i havent even come across some of the notation in them! Does STEP imply that people doing these read around what they've been taught?

I'm a (long time ago) Cambridge graduate, so you can assume I've read around rather a lot more than most people. I am very rusty though!

I would say STEP I, STEP II only really require what's on the syllabus (though bear in mind I don't know the current syllabi terribly well!), although you do have to be very familiar with everything, particularly all the trig identities etc.

On the other hand, it seems STEP III has a nasty habit of taking a problem that comes up in university maths and rephrasing it, which means those of us with that background tend to "push" the problem towards something we're familiar with, and it is much easier for someone with a lot of background knowledge. (I think this is less true in recent years for STEP III).

As far as notation goes, post a question here and we'll try and explain it. The only "esoteric" notation I've noticed was my fault! I ended up using the

\prod

symbol in one question - I wouldn't have done so writing by hand, but it seemed easier given forum constraints.

But

\prod

this is basically the same as

\sum

only for products. e.g.

\prod_{i=1}^5 i = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 120

\prod_{i=1}^N i = N!

As far as finding it difficult - only being able to do 1 or 2 questions a paper is completely normal at first (in fact, I'd say it's pretty good going). Practice does make a huge difference.

Reply 55

Well. I think I am guilty of reading around the subject, where reading around generally means reading through interesting bits off wikipedia and mathworld, and using those to read about terminology and methods and other stuff that I encounter here on these forums or elsewhere which I don't know. I feel I have the basic skills to solve most of the pure questions in STEP I and II (but usually for statistics at least I need to look stuff up), whereas STEP III is more about "look through the paper", "eliminate the questions that are on topics I obviously don't know", "eliminate the questions that obviously need methods that I don't know" and then try to solve the 3 or 4 remaining. Fortunately, I only have to do STEP I and STEP II for my Cambridge offer.

The issue I'm having is more about time, when working at home, I can solve most of the questions after a while, but I'm not confident that I can solve and write solutions to four of them in three hours. But I guess that's where practice comes in...

And oh, I'm definitely guilty of abusing the

\prod

symbol (in my solution to Q3 STEP II in the 1998 thread), though I was actually taught the symbol in school (only very very briefly in passing when being introduced to the

\sum

symbol, but taught the symbol nevertheless). Then, again, I realise that you'll most probably have to read my solution on that question through a lot of times before you understand what those product symbols actually mean, but I mean, meh, who cares? :p:

Using them removes any sort of ambiguity, yes, it removes any sort of intelligibility as well, but you have to make sacrifices, don't you...

Reply 56